Q. 254.2( 13 Votes )

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Answer :


As cosθ+cos2 θ=1
cos θ =1-cos2 θ
cos θ =sin2 θ ….(i)

Using(a + b)3 =a3+b3+3(a + b)
Now,sin12 θ+3sin10 θ+3sin8 θ+sin6 θ+2sin4 θ+2sin2 θ-2= (sin4 θ)3+sin4 θ-sin2 θ[sin4 θ+sin2 θ](sin2 θ)3+2(sin2 θ)2+2sin2 θ-2
=(sin4 θ+sin2 θ)3+2(cos θ)2+2cosθ-2.
=((sin2 θ)2+sin2 θ)3 +2cos2 θ+2cosθ-2

From (1),
=(cos2 θ +sin2 θ)3+2cos2 θ+2cosθ-2

As sin2θ + cos2θ = 1
= 1 +2cos2 θ+2sin2 θ-2
= 1 +2(cos2 θ+sin2 θ)-2


Hence proved.

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